SOLUTION: A clothing manufacturer has 100 m of sill and 180 of wool. To make a suite it requires 2m of silk adn 3m of wool, and to make a dress 1m of silk and 2m of wool. If the profit on th

Question 140853: A clothing manufacturer has 100 m of sill and 180 of wool. To make a suite it requires 2m of silk adn 3m of wool, and to make a dress 1m of silk and 2m of wool. If the profit on the suit is $108 and the profit on the dress is $60, how many suits and dressed should be madeto maximize the profit?

I started something but it's wrong. Please help!
2x+1y=108
x+2y= 60
Profit = 108x+60y

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A clothing manufacturer has 100 m of silk and 180 m of wool.
To make a suit it requires 2m of silk and 3m of wool,
to make a dress 1m of silk and 2m of wool.
If the profit on the suit is $108 and the profit on the dress is $60,
how many suits and dresses should be made to maximize the profit?
-------------------------
Let # of suits be "s"; Let # of dresses be "d":
Profit = 108s + 60d
--------------------
Silk: 2s + d <= 100
Wool: 3s +2d <= 180
----------------------
Solve both for d and graph them:
d <= -2s + 100
d <= (-3/2)s + 90
-----------------------

-----------------------
Find the coordinates of the enclosed polygon and test the
coordinates in the Profit equation to determine the maximum.
-----------------------
Coordinates: (20,60), (0,0) , (0,90), (50,0)
-----------------------
Determining maximum Profit:
Profit = 108s + 60d
(20,60): P = 108*20 + 60*60 = 5760
(0,90): P = 90*60 = 5400
(50,0): p = 50*108 = 5400
------------------------------
Maximum profit comes when 20 dresses and 60 suits are produced
================================
Cheers,
Stan H.
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